Round 84552 to the nearest thousands

g The downtime per day for a computing facility has mean 4 hours and standard deviation 0.9 hour. What assumptions must be true

Sedaia [141]

Answer:

To obtain a valid approximation for probabilities about the average daily downtime, either the underlying distribution(of the downtime per day for a computing facility) must be normal, or the sample size must be of 30 or more.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

To obtain a valid approximation for probabilities about the average daily downtime, either the underlying distribution(of the downtime per day for a computing facility) must be normal, or the sample size must be of 30 or more.

7 0

3 years ago

22.5 + 7(n−3.4) simplified

NARA [144]

-1.3+7n is the answer

7 0

3 years ago

Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 a

krok68 [10]

Answer:

Jessie scored higher than Reagan.

Step-by-step explanation:

We are given that Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 and standard deviation 100.

Jessie scored 30 on the ACT. The distribution of ACT scores in a reference population is normally distributed with mean 17 and standard deviation 5.

For finding who performed better on the standardized exams, we have to calculate the z-scores for both people.

1. <u>Finding z-score for Reagan;</u>

Let X = distribution of SAT scores

SO, X ~ Normal( $\mu=1000, \sigma^{2}=100^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 1000

$\sigma$ = standard deviation = 100

Now, Reagan scored 1140 on the SAT, that is;

z-score = $\frac{1140-1000}{100}$ = 1.4

2. <u>Finding z-score for Jessie;</u>

Let X = distribution of ACT scores

SO, X ~ Normal( $\mu=17, \sigma^{2}=5^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 17

$\sigma$ = standard deviation = 5

Now, Jessie scored 30 on the ACT, that is;

z-score = $\frac{30-17}{5}$ = 2.6

This means that Jessie scored higher than Reagan because Jessie's standardized score was 2.6, which is 2.6 standard deviations above the mean and Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean.

6 0

3 years ago

Which of the following is the solution to the compound inequality below?

Minchanka [31]

Answer:

b. x ≥-2 or x ≤-6

7 0

2 years ago

The answer to the picture please

Vinil7 [7]

the required ans is 3√b+b/3b

8 0

3 years ago

Round 84552 to the nearest thousands​

Round 84552 to the nearest thousands